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We will discuss here about the law of sines or the sine rule which is required for solving the problems on triangle.
In any triangle the sides of a triangle are proportional to the sines of the angles opposite to them.
That is in any triangle ABC,
asinA = bsinB = csinC
Proof:
Let ABC be a triangle.
Now will derive the three different cases:
Case I: Acute angled triangle (three angles are acute): The triangle ABC is acute-angled.
Now, draw AD from A which is perpendicular to BC. Clearly, D lies on BC
Now from the triangle ABD, we have,
sin B = AD/AB
β sin B = AD/c, [Since, AB = c]
β AD= c sin B β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦. (1)
Again from the triangle ACD we have,
sin C = AD/AC
β sin C = AD/b, [Since, AC = b]
β AD = b sin C ...β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦.. (2)
Now, from (1) and (2) we get,
c sin B = b sin C
β b/sin B = c/sin cβ¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦.(3)
Similarly, if we draw a perpendicular to AC from B, we will get
a/sin A = c/sin cβ¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦.(4)
Therefore, from (3) and (4) we get,
asinA = bsinB = csinC
Case II: Obtuse angled triangle (one angle is obtuse): The triangle ABC is obtuse angled.
Now, draw AD from A which is perpendicular to produced BC. Clearly, D lies on produced BC.
Now from the triangle ABD, we have,
sin β ABD = AD/AB
β sin (180 - B) = AD/c, [Since β ABD = 180 - B and AB = c]
β sin B = AD/c, [Since sin (180 - ΞΈ) = sin ΞΈ]
β AD = c sin B β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦. (5)
Again, from the triangle ACD, we have,
sin C = AD/AC
β sin C = AD/b, [Since, AC = b]
β AD = b sin C β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦. (6)
Now, from (5) and (6) we get,
c sin B = b sin C
b/sin B = c/sin C β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦. (7)
Similarly, if we draw a perpendicular to AC from B, we will get
a/sin A = b/sin B β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦. (8)
Therefore, from (7) and (8) we get,
asinA = bsinB = csinC
Case III: Right angled triangle (one angle is right angle): The triangle ABC is right angled. The angle C is a right angle.
Now from triangle ABC, we have,
sin C = sin Ο/2
β sin C = 1, [Since, sin Ο/2 = 1], β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦. (9)
sin A = BC/AB
β sin A = a/c, [Since, BC = a and AB = c]
β c = a/sin A β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦. (10)
and sin B = AC/AB
β sin B = b/c, [Since, AC = b and AB = c]
β c = b/sin B β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦. (11)
Now from (10) and (11) we get,
a/sin A = b/sin B = c
β a/sin A = b/sin B = c/1
Now from (9) we get,
β asinA = bsinB = csinC
Therefore, from all three cases, we get,
asinA = bsinB = csinC. Proved.
Note:
1. The sine rule or the law of sines can be expressed as
sinAa = sinBb = sinCc
2. The sine rule or the law of sines is a very useful rule to
express sides of a triangle in terms of the sines of angles and vice-versa in
the following manner.
We have asinA = bsinB = csinC = k1 (say)
β a = k1 sin A, b = k1 sin B and c = k1 sin C
Similarly, sin A/a = sin B/b = sin C/c = k2 (say)
β sin A = k2 a, sin B = k2 b and sin C = k2 c
Solved problem using the law of sines:
The triangle ABC is isosceles; if β A = 108Β°, find the value of a : b.
Solution:
Since the triangle ABC is isosceles and A = 108Β°, A + B + C = 180Β°, hence it is evident that B = C.
Now, B + C = 180Β° - A = 180Β° - 108Β°
β 2B = 72Β° [Since, C = B]
β B = 36Β°
Again, we have, asinA = bsinB
Therefore, ab = sinAsinB = sin108Β°sin36Β° = cos18Β°sin36Β°
Now, cos 18Β° = β1βsin218Β°
= β1β(β5β14)2
= ΒΌβ10+2β5
and sin 36Β° = β1βcos236Β°
= β1β(β5+14)2
= ΒΌβ10β2β5
Therefore, a/b = 14β10+2β514β10β2β5
= β10+2β5β10β2β5
= β(10+2β5)2102β(2β5)2
= 10+2β5β80
β ab = 2β5(β5+1)4β5
β ab = β5+12
Therefore, a : b = (β5 + 1) : 2
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